research article ultra-high-speed travelling wave...

Research ArticleUltra-High-Speed Travelling Wave Protection of TransmissionLine Using Polarity Comparison Principle Based on EmpiricalMode Decomposition

Dong Wang12 Houlei Gao12 Sibei Luo12 and Guibin Zou12

1Key Laboratory of Power System Intelligent Dispatch and Control ofMinistry of Education Shandong University Jinan 250061 China2School of Electrical Engineering Shandong University Jinan 250061 China

Correspondence should be addressed to Houlei Gao houleigsdueducn

Received 30 July 2015 Revised 8 September 2015 Accepted 9 September 2015

Academic Editor Peter Dabnichki

Copyright copy 2015 Dong Wang et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The traditional polarity comparison based travelling wave protection using the initial wave information is affected by initial faultangle bus structure and external fault And the relationship between the magnitude and polarity of travelling wave is ignoredBecause of the protection tripping and malfunction the further application of this protection principle is affected Therefore thispaper presents an ultra-high-speed travelling wave protection using integral based polarity comparison principle After empiricalmode decomposition of the original travelling wave the first-order intrinsic mode function is used as protection object Based onthe relationship between the magnitude and polarity of travelling wave this paper demonstrates the feasibility of using travellingwave magnitude which contains polar information as direction criterion And the paper integrates the direction criterion in aperiod after fault to avoid wave head detection failure Through PSCAD simulation with the typical 500 kV transmission systemthe reliability and sensitivity of travelling wave protection were verified under different factorsrsquo affection

1 Introduction

According to the protection principle travelling wave protec-tion methods include travelling wave differential protectiontravelling wave distance protection travelling wave ampli-tude comparison protection and travelling wave polaritycomparison protection [1 2]

Travelling wave differential protection principle is simpleand clear But travelling wave has attenuation characteristicThere may be large unbalance current in the transmissionline to cause wrong operation And it is also affected bythe bus structure [3ndash5] Travelling wave distance protectioncannot protect the whole line and does not have directiondiscrimination ability Same with differential protection itis affected by bus structure too [6ndash9] Travelling waveamplitude comparison protection principle has improvedmuch compared to other protection principles But it isaffected by the bus structures fault inception angles and dif-ferent thresholds tooThe traditional travelling wave polaritycomparison protection principle has a lot of advantages high

operation speed clear direction discrimination and simpleprotection principle But it is also affected by some thingsfault initial angles different bus structures different faultlocations and even threshold If those disadvantages can beovercome a new travelling wave protection principle can begot [10ndash12]

The traditional travelling wave protection principlersquosapplication is limited by the transformer technology Thetraditional current transformer (CT) and voltage transformer(VT) cannot transfer the travelling wave signal correctlyCurrently Rogowski based electronic current transformer (R-ECT) and capacitive divider electronic voltage transformer(C-EVT) have been able to transfer current travelling waveand voltage travelling wave accurately And the output of C-EVT and R-ECT is the differential signal of the input Byintegration circuit the original signal can be regained exactlySo there is no transformer technology limit in the travellingwave protection principle anymore [13ndash16]

This paper compares the polarity relationship betweenvoltage travelling wave and current travelling wave with

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 195170 9 pageshttpdxdoiorg1011552015195170

2 Mathematical Problems in Engineering

Table 1 Analysis of travelling wave polarity comparison protection

Fault location Superimposed voltagepolarity

M side N side Schematic of fault superimposedstate circuitVoltage Current Voltage Current

Internal fault +minus

+minus

minus

++minus

minus

+ ++

M N

minus

minus

External fault +minus

+minus

minus

++minus

+minus

M N

++minus

minus


+minus

+minus

+minus

minus

+

M N

++ minusminus

different fault directions Combined with empirical modedecomposition algorithm (EMD) the new travelling wavepolarity comparison protection principle based on the inte-gration of amplitude is derived It not only uses the initialtravelling wave but also uses the travelling wave after faulthappens So it is a reliable protection principle with obviousdirection discrimination By the way this new protectionprinciple is not affected by different initial angles differentgrounding resistance different bus structures and differentfault locations To verify the characteristics of the new prin-ciple a simulation based on PSCADEMTDC is carried onAnd the PSCAD simulation proved that this new protectionprinciple has the characteristics mentioned above indeed

The high operation speed is a very important advantagefor travelling wave protection Considering the differentparts of the new travelling waversquos operation time the newprotection principle can determine if the fault is internal orexternal in 5ms Then it can send the signal to breaker tooperate So it can be called an ultra-high-speed travellingwave protection

2 Traditional Travelling Wave PolarityComparison Protection

The direction element of the protection principle is a polaritycomparison relay It detects the initial voltage travelling waveand current travelling wave as comparison objects Whenthe voltage travelling wave and current travelling wave haveopposite polarities of both sides the internal fault can bedetermined When the voltage travelling wave and currenttravelling wave have same polarities of any side the externalfault can be determinedThe schematic of protection is shownin Table 1 (Superimposed voltage in the table appears at themoment that fault happens It has same value and oppositepolarity with the voltage on the transmission line just beforethe fault moment And it is the voltage source in the circuit ofthe table indeed)

3 Empirical Mode Decomposition

The empirical mode decomposition algorithm can distin-guish the different scale fluctuations or trends in the signalgradually And the result of EMD is a series of differentcharacteristic scales data called intrinsic mode functions(IMF) [17ndash19]

The result of EMD can be described as

119878 (119905) =

119873

sum

119894=1

IMF119894 (119905) + 119877 (119905)

(1)

where 119878(119905) is the EMD result IMF119894(119905) is 119894 order intrinsicmode

function and 119877(119905) is trends signalIntrinsic mode function is a single component signal and

it must meet the following two conditions (1) differencebetween the number of extreme points and zero crossingpoints is not more than one over the entire length of thesignal (2) the envelope of IMF is symmetry about time axis

The processes of EMD are described as the followingsteps

(1) Find all the maxima of the original signal 119878(119905) Thenthe maxima envelope 119864

+(119905) can be calculated based on cubic

spline interpolation Similarly the minimum envelope 119864

minus(119905)

can also be calculated Then the average envelope 119872(119905) canbe defined as

119872(119905) =

119864

+ (119905) + 119864

minus (119905)

2

(2)

(2) Let 119878(119905) be minus119872(119905) to get a new signal11986711(119905)

119867

1

1(119905) = 119878 (119905) minus 119872 (119905)

(3)

Then check if the following condition can be met

sum[119867

119896

1(119905) minus 119867

119896minus1

1(119905)]

2

sum[119867

119896minus1

1(119905)]

2le 120576

(4)

Mathematical Problems in Engineering 3

where 119896 is the cycle number and the value of 120576 is between 02and 03 This paper selects 03

If (4) cannot be met return to step (1)If (4) can be met after 119896 cycles then IMF

1(119905) can be

defined as

IMF1 (

119905) = 119867

119896

1(119905)

(5)

(3) Let original signal be minus IMF1(119905) to get a residual

signal 119877(119905) as

119877 (119905) = 119878 (119905) minus IMF1 (

119905) (6)

Repeat steps (1) to (3) to get another intrinsic modefunction IMF

2(119905) Repeat steps (1) to (3) until residual signal

119877(119905) is small enough or monotonic functionBased on EMD the first IMF of voltage travelling wave

and current travelling wave can be calculated as Figure 1The first figure and second figure are the voltage and currenttravellingwave before EMD respectivelyThe third figure andfourth figure are the voltage and current travelling wave afterEMD respectively As we can see the similarity of the voltageand current travelling wave using EMD is more obvious thanbefore So it is more convenient to construct a protectionprinciple using intrinsic mode function

4 New Travelling Wave PolarityComparison Protection

41 Derivation of Direction Criterion As shown in Figure 2fault component voltage appears between the fault location119865 in transmission line and the earth at time 119905 = 0Then the transmission line will be charged After a shorttime Δ119905 a short length of transmission line Δ119909 is chargedto Δ119876 = 119862119906

0Δ119909 (119862 is the capacitance value per unit

length of transmission line) An electrical field 119864 will appearsurrounding this short transmission line And the flow ofcurrent will form a magnetic field around the line If Δ119909 issmall enough the current 119894

0can be described as

119894

0= limΔ119909rarr0

Δ119876

Δ119905

= limΔ119909rarr0

119862119906

0Δ119909

Δ119905

= 119862119906

0V (7)

where V is the speed of the wave and119862 is the capacitance valueper unit length of transmission line

Now the magnetic flux around Δ119909 is ΔΦ = 119871119894Δ119909According to the law of electromagnetic induction theelectromotive force is described as

119864 = limΔ119909rarr0

ΔΦ

Δ119905

= limΔ119909rarr0

119871119862119906

0VΔ119909

Δ119905

= 119871119862119906

0V2 (8)

where V is the speed of the wave 119862 is the capacitance valueper unit length of transmission line and 119871 is the inductancevalue per unit length of transmission line

Because the voltage on capacitance cannot change sud-denly and Δ119909 is small enough 119864 equals voltage 119906

0 Then the

wave speed will be

V =

1

radic

119871119862

(9)

minus500

50

um (k

V)

im (k

A)

umim

f (kV

)im

imf (

kA)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

minus04minus02

0

minus010

01

minus010

01

Figure 1 The comparison of travelling wave and travelling waveafter EMD

Bus

Bus F

F

R1R2

HH

E E

i0i0

C2C9984002

u0

Δx

C1

fminus

f+

f0

f998400+

Figure 2 Equivalent circuit of single phase line

Take (9) to (7) to get

119906

0

119894

0

=

radic

119871

119862

(10)

Aswe can see the ratio of voltage and current is a constantvalue called wave impedance

Define voltage amplitude conditioning factor 119896

119896 =

|119894|

|119906|

(11)

And the value of 119896 is approximately equal toradic119862119871Considering the initial polarity of voltage and current

travelling wave define a factor 120582 to identify the fault direc-tion

120582 =

int

119905119865+119889

119905=119905119865

119896119906 (119905) 119894 (119905) 119889119905

int

119905119865+119889

119905=119905119865

119894

2(119905) 119889119905

(12)


where 119905

119865is the arrival point of the travelling wave 119889 is the

length of the integration time and 119896 is the voltage amplitudeconditioning factor defined above

The discretization of (12) can be described as

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

(13)

Considering different fault directions the value of 120582 canbe calculated

(1) If the fault happens as Figure 2 shows it will beforward fault type for R1 Assume the transmission line islosslessWhen the travelling wave arrived the travelling wavesignal of R1 will be

119906 = 119906

++ 119906

minus= (1 + 119896

119906119891) 119906

minus

119894 = 119894

++ 119894

minus= (1 + 119896

119894119891) 119894

minus

(14)

where 119906 is the voltage travelling wave 119894 is the currenttravelling wave 119906

+is the forward voltage travelling wave 119906

minus

is the reverse voltage travelling wave 119894+is the forward current

travelling wave 119894minusis the reverse current travelling wave 119896

119906119891

is the voltage reflection coefficient at the bus and 119896

119894119891is the

current reflection coefficient at the bus And the inequalityrelationship will be 0 le |119896

119906119891| |119896

119894119891| le 1

Now the amplitude conditioning factor can be calculated

119896 =

|119894 (119905)|

|119906 (119905)|

=

(1 + 119896

119894119891)

(1 + 119896

119906119891)

1003816

1003816

1003816

1003816

119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119906

minus (119905)

1003816

1003816

1003816

1003816

(15)

Because it is a forward direction fault for R1 the reversevoltage and current travelling wave have different polaritiesSo (15) can be simplified as

119896 = minus

(1 + 119896

119894119891)

(1 + 119896

119906119891)

119894

minus (119905)

119906

minus (119905)

(16)

Take (14) and (16) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= minus1 (17)

Aswe can see the value of120582 is a constant numberminus1 whenit is a forward direction fault And it is not affected by thereflection coefficient and the construction of the bus

(2) If the fault happens as Figure 2 shows it will be reversefault type for R2 Assume the transmission line is losslessWhen the travelling wave arrived the travelling wave signalof R2 will be

119906 = 119906

1015840

+= 119896

119906119911119906

minus

119894 = 119894

1015840

+= 119896

119894119911119894

minus

(18)


+is the forward voltage travelling wave at

R2 1198941015840+is the forward current travelling wave at R2 119906

minusis the

minus1 +10

Forward fault Reverse faultminus1 le 120582 lt 0 0 lt 120582 le 1

Figure 3 Fault direction discrimination schematic

reverse voltage travelling wave at R1 119894minusis the reverse current

travelling wave at R1 119896119906119911

is the voltage refractive coefficientat bus and 119896

119894119911is the current refractive coefficient at bus And

there is an inequality relationship 119896

119906119911 119896

119894119911ge 0


119896 =

|119894 (119905)|

|119906 (119905)|

=

1003816

1003816

1003816

1003816

119896

119894119911119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119896

119906119911119906

minus (119905)

1003816

1003816

1003816

1003816

(19)

Because it is a reverse direction fault for R2 and also aforward direction fault for R1 the reverse voltage and currenttravelling wave have different polarities And the forwarddirection of R2 is opposite to R1 So (19) can be simplified as

119896 =

|119894 (119905)|

|119906 (119905)|

=

119896

119894119911

119896

119906119911

119894

minus (119905)

119906

minus (119905)

(20)

Take (18) and (20) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= 1 (21)

As we can see the value of 120582 is a constant number 1 whenit is a reverse direction fault And it is not affected by thereflection coefficient and the construction of the bus

Taking a variety of errors in the actual system intoaccount the fault direction discrimination schematic isshown in Figure 3 When forward fault happens the value of120582 is less than zero When reverse fault happens the value of120582 is greater than zero If two fault direction discriminationresults of both ends are forward fault an internal fault canbe determined If one of the fault direction discriminationresults of both ends is reverse fault an external fault can bedetermined

42 Protection Scheme The protection scheme is shownin Figure 4 First of all the three-phase voltage and cur-rent should be decoupled using Clark transformation Thenamplitude conditioning factor defined above can be calcu-lated point by point in 119889 length of time After that the value of120582 can be calculated Because the other end of the transmissionline needs the value of 120582 to identify the fault section the valueof 120582 should send to another end though fiber path Then thevalue of factor 120582 can be checked to identify the external faulttype If it is a forward fault for the relay the value of factor 1205821015840from another end will be received and checked to identify theexternal fault If 120582 and 120582

1015840 are both less than zero an internalfault can be determined At last the breaker will clear thetransmission line fault

5 Simulation Analysis

51 Simulation Model in PSCAD The 500 kV power trans-mission system is constructed in PSCADEMTDC as shown


Travelling wave data

Clark transformation

Yes

Yes

No

No

another end

from another end

Return

Internal fault

External fault

Return

External fault

Breaker action

of time

Return

point in d length of timeCalculate k point by

Calculate 120582 in d length

Send the value of 120582 to

0 lt 120582 le 1

minus1 le 120582998400 lt 0

Receive the value of 120582998400

Figure 4 Flow chart of travelling wave protection

Table 2 Transmission line parameters

119877 119883 119866 119861

(Ωkm) (Ωkm) (Skm) (Skm)Positive sequence 001798 029278 1 times 108 393905 times 10minus6

Negative sequence 001798 029278 1 times 108 393905 times 10minus6

Zero sequences 028662 108210 1 times 108 243767 times 10minus6

in Figure 5 The system includes three transmission lineswhose lengths are 100 km 200 km and 100 km respectivelyR1 and R2 are two relays on the middle line Now thenew travelling wave polarity comparison protection can bestudied by different fault locations and different fault types

The transmission line uses frequency-dependent modeland has uniform transpositionThe transmission line param-eters for per km length are shown in Table 2 The bus straycapacitance to ground is set to 119862

119878= 001 120583F Taking the past

studies into account the sampling rate is set to 1MHz Theintegration time 119889 is 01ms

52 Typical Fault Examples In order to verify the protectionprinciplersquos operating characteristics A phase to ground fault

M NR1 R2F1F2 F3

CSCSCSCS

L2 = 100 km L1 = 200 km L3 = 100 km

Figure 5 Model of 500 kV power transmission system

minus005

0

005

ku

(kV

)

minus005

0

005

i (kA

)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

Figure 6 Comparison chart of 119896119906 and 119894 of M side

is set located at F3The initial fault angle is 45∘ and the groundresistance is 50Ω Using empirical mode decompositionalgorithm the first-order intrinsic mode function of voltageand current travelling wave of both sides can be calculated

Take 119896119906 and 119894 data in Figure 6 to (13) to calculate the 120582 =

minus08710 Then the forward fault of R1 can be determinedTake 119896119906 and 119894 data in Figure 7 to (13) to calculate the 120582 =

09811 Then the reverse fault of R2 can be determinedAs we can see the fault discrimination results of R1 and

R2 are corrected Taking the protection scheme of Figure 4into account an external fault type can be determined

53 Relater Factors

531 Different Fault Location Based on some different faultlocations at F1 andF2 the fault discrimination factor120582of bothM and N side is calculated

Table 3 is the simulation results for different fault loca-tions The fault distance in the table is from the bus of Mside to fault location As can be seen the fault principle basedon EMD can identify fault direction correctly Even at thebeginning or end of the transmission line it can still identifyfault direction correctly

532 Grounding Resistance Based on some differentgrounding resistance at F1 (100 km away from the bus of Mside) and F2 (10 km away from the bus of M side) the faultdiscrimination factor 120582 of both M and N side is calculated

Table 4 is the simulation results for different groundingresistance As can be seen the fault principle based onEMD can identify fault direction correctly With the increas-ing of grounding resistance protectionrsquos sensitivity will notchange

6 Mathematical Problems in Engineeringku

(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01

Figure 7 Comparison chart of 119896119906 and 119894 of N side

533 Fault Initial Angle Based on some different fault initialangel at F1 (100 km away from the bus of M side) and F2(10 km away from the bus of M side) the fault discriminationfactor 120582 of both M and N side is calculated

Table 5 is the simulation results for different fault initialangles As can be seen the fault principle based on EMDcan identify fault direction correctly Even with small faultangles it can still identify fault direction correctly And withthe decreasing of fault initial angle protectionrsquos sensitivitywillreduce slowly

534 Different Fault Types Based on some different faulttypes at F1 (100 km away from the bus of M side) and F2(10 km away from the bus of M side) the fault discriminationfactor 120582 of both M and N side is calculated

Table 6 is the simulation results for different fault typesAs can be seen the fault principle based on EMD can identifyfault direction correctly

535 Sampling Rate Based on some different sampling rateand AG fault at F1 (100 km away from the bus of M side)and F2 (10 km away from the bus of M side) the faultdiscrimination factor 120582 of both M and N side is calculated

Table 7 is the simulation results for different samplingrate As can be seen the fault principle based on EMD canidentify fault direction correctly with the change of samplingrate

536 Bus Structure Traditional travelling wave protectionprinciple is affected by the number of transmission linesconnected to the bus To verify the new EMD basedprotection principle a new power transmission system isconstructed in PSCAD as Figure 8

Table 8 is the simulation results for different fault loca-tions As can be seen the fault principle based on EMD canidentify fault direction correctly with different bus structure

6 Operation Time of Protection

The operation time of the travelling wave protection usingpolarity comparison principle based on EMD includes threeparts algorithm time detection time and propagation timeThis new travelling wave protection principle can determineif the fault is inside or outside of the protection region in

M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3

Figure 8 Model of 500 kV power transmission system with differ-ent bus structures

M NR1 R2F

CS CS

L2L1

tftm

tn

Figure 9 Schematic diagram of detection time

5ms Then it can send the signal to breaker to operate So itcan be called ultra-high-speed travellingwave protectionThefollowing is the introduction and analysis of the three parts

61 Algorithm Time Algorithm time includes two parts theintegration time and calculation time of the principle In thispaper integration time length (the factor 119889 in (12) and (13))is 01ms Considering the computing power of the protectionunit now the calculation time of algorithm is not longer than05ms So the algorithm time is not longer than 1ms

62 Detection Time Detection time is the time difference oftwo sidesrsquo travellingwave arrival point Aswe can see the faultmay happen everywhere in the transmission line Then thearrival times of two sides are different except that the faulthappens in the middle of the line As a protection principlewhich needs two sidesrsquo information to decide the operationof breaker the time difference will delay the operation timeAs shown in Figure 9 119905

119891 119905119898 and 119905

119899are the fault timeM sidersquos

arrival time and N sidersquos arrival time respectively And thetime difference can be described as

Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)

And V is the travelling wave speedBecause the transmission line is generally several hun-

dred kilometers this time is obviously not longer than 2ms

63 Propagation Time After the direction discrimination ofone side as shown in Figure 10 the value of 120582 should transferto another side Propagation time is the time from one side toanother side As the length of transmission line is generally


Table 3 Simulation results for different fault locations

Fault location Fault distancekm M side N side Results120582 Direction 120582 Direction

F110 minus08160 Forward minus05732 Forward Internal100 minus08763 Forward minus04071 Forward Internal190 minus09378 Forward minus07965 Forward Internal

F210 09972 Reverse minus09603 Forward External50 09980 Reverse minus03843 Forward External90 09972 Reverse minus09603 Forward External

Table 4 Simulation results for different grounding resistance

Fault location Grounding resistanceΩ M side N side Results120582 Direction 120582 Direction



Table 5 Simulation results for different fault angles

Fault location Initial angle∘ M side N side Results120582 Direction 120582 Direction



Table 6 Simulation results for different fault types

Fault location Fault type M side N side Results120582 Direction 120582 Direction

F1

AG minus04944 Forward minus08818 Forward InternalAC minus09600 Forward minus09602 Forward InternalABG minus04471 Forward minus08804 Forward InternalABCG minus09664 Forward minus07452 Forward Internal

F2

AG 09972 Reverse minus09700 Forward ExternalAC 09991 Reverse minus06022 Forward ExternalABG 08275 Reverse minus04659 Forward ExternalABCG 09997 Reverse minus09586 Forward External

Table 7 Simulation results for different sampling rate

Fault location Sampling rateHz M side N side Results120582 Direction 120582 Direction

F1100 k minus09794 Forward minus06816 Forward Internal500 k minus07345 Forward minus09637 Forward Internal1M minus05018 Forward minus08825 Forward Internal

F2100 k 09751 Reverse minus09807 Forward External500 k 09994 Reverse minus09055 Forward External1M 09979 Reverse minus09684 Forward External



Fault location M side N side Results120582 Direction 120582 Direction

F1 minus09731 Forward minus09603 Forward InternalF2 09980 Reverse minus09779 Forward ExternalF3 minus09718 Forward 09992 Reverse External

M NR1 R2F

L

L2L1

CS CS

120582m 120582n

Figure 10 Schematic diagram of propagation time

several hundred kilometers propagation time is no longerthan 2ms

7 Conclusion

Comparing with the traditional polarity comparison trav-elling wave protection the new travelling wave protectioncombines the relationship between amplitude and polarityBased on empirical mode decomposition the derivationof the direction criterion is finished And this protectioncriterion not only uses the initial travelling wave front butalso uses short timersquos (01ms in the paper) travelling waveinformation after the initial travelling wave front Throughthe integration of travelling wave it can avoid the failure ofthe travellingwaversquos detection So it can increase the reliabilityof the protection principle

To verify the new protection principle a simulationbased on PSCAD is carried on Taking the simulation resultsinto account this new protection principle is not affectedby different fault locations different fault types differentinitial angels different grounding resistance and differentbus structures So it is a reliable travelling wave protection

Operation speed is an important advantage for travellingwave protection Because the new protection principle cansend the operation signal to breaker in 5ms it can be calledultra-high-speed travelling wave protection

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the republication of this paper

Acknowledgment

This work was supported in part by the National NaturalScience Foundation of China (51177094 51277114)

References

[1] G Zou andH Gao ldquoA traveling-wave-based amplitude integralbusbar protection techniquerdquo IEEE Transactions on PowerDelivery vol 27 no 2 pp 602ndash609 2012

[2] W Chen O P Malik X Yin D Chen and Z Zhang ldquoStudyof wavelet-based ultra high speed directional transmission lineprotectionrdquo IEEE Transactions on Power Delivery vol 18 no 4pp 1134ndash1139 2003

[3] G Zou S Song C Xu D Liu and H Gao ldquoFast busbar protec-tion based on waveform integral of directional traveling wavesrdquoProceedings of the Chinese Society of Electrical Engineering vol34 no 31 pp 5677ndash5684 2014

[4] Z Li andHHua ldquoTravelingwave protection based onwide areatravelling wave informationrdquo Proceedings of the CSEE vol 34pp 6238ndash6245 2014

[5] J-D Duan B-H Zhang S-B Luo and Y Zhou ldquoTransient-based ultra-high-speed directional protection using wavelettransforms for EHV transmission linesrdquo in Proceedings ofthe IEEEPES Transmission and Distribution Conference andExhibition Dalian China August 2005

[6] Q H Wu J F Zhang and D J Zhang ldquoUltra-high-speeddirectional protection of transmission lines usingmathematicalmorphologyrdquo IEEE Transactions on Power Delivery vol 18 no4 pp 1127ndash1133 2003

[7] G Zou Study on internal based travelling wave directional pro-tection for transmission line [PhD thesis] Shandong UniversityJinan China 2009

[8] M Ohrstrom andL Soder ldquoFast protection of strong power sys-tems with fault current limiters and PLL-aided fault detectionrdquoIEEE Transactions on Power Delivery vol 26 no 3 pp 1538ndash1544 2011

[9] G B Zou and H L Gao ldquoExtra high speed hybrid protectionscheme for high voltage transmission linerdquo International Jour-nal of Electrical PowerampEnergy Systems vol 63 pp 83ndash90 2014

[10] H Ha and Y Yu ldquoNovel scheme of travelling wave baseddifferential protection for bipolar HVDC transmission linesrdquoin Proceedings of the International Conference on Power SystemTechnology (POWERCON rsquo10) pp 1ndash6 IEEE Hangzhou ChinaOctober 2010

[11] J D Duan and B H Zhang ldquoStudy of starting algorithm usingtravelling wavesrdquo Proceedings of the Chinese Society for ElectricalEngineering vol 24 no 9 pp 30ndash36 2004

[12] G B Zou and H L Gao ldquoAlgorithm for ultra high speedtravelling wave protection with accurate fault locationrdquo inProceedings of the IEEE Power and Energy Society GeneralMeeting pp 20ndash24 Pittsburgh Pa USA July 2008

[13] M Marracci B Tellini C Zappacosta and G Robles ldquoCriticalparameters for mutual inductance between rogowski coil andprimary conductorrdquo IEEE Transactions on Instrumentation andMeasurement vol 60 no 2 pp 625ndash632 2011

[14] W B Li C X Mao and J M Lu ldquoStudy of Rogowski coilsfor measuring pulse currents of the high-power laser sourcerdquoInternational Journal of Power and Energy Systems vol 2 pp96ndash101 2005

[15] Y Liu F C Lin Q Zhang and H Zhong ldquoDesign andconstruction of a Rogowski Coil for measuring wide pulsedcurrentrdquo IEEE Sensors Journal vol 11 no 1 pp 123ndash130 2011

[16] H Gao and R Yang ldquoAnalysis and test for electronic voltagetransducerrsquos transfer characteristicsrdquo Journal of Beijing JiaotongUniversity vol 38 no 5 pp 114ndash118 2014


[17] N Huang and M Wu ldquoA confidence limit for the empiricalmode decomposition andHilbert spectral analysisrdquo Proceedingsof the Royal Society A Mathematical Physical and EngineeringSciences vol 459 no 2037 pp 2317ndash2345 2003

[18] N E Huang Z Shen S R Long et al ldquoThe empiricalmode decomposition and the Hilbert spectrum for nonlinearand non-stationary time series analysisrdquo The Royal Societyof London ProceedingsmdashSeries A Mathematical Physical andEngineering Sciences vol 454 no 1971 pp 903ndash995 1998

[19] Z Wu and N E Huang ldquoA study of the characteristics ofwhite noise using the empirical mode decomposition methodrdquoProceedings of the Royal Society A Mathematical Physical andEngineering Sciences vol 460 no 2046 pp 1597ndash1611 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Table 1 Analysis of travelling wave polarity comparison protection

Fault location Superimposed voltagepolarity

M side N side Schematic of fault superimposedstate circuitVoltage Current Voltage Current

Internal fault +minus

+minus

minus

++minus

minus

+ ++

M N

minus

minus


+minus

minus

++minus

+minus

M N

++minus

minus


+minus

+minus

+minus

minus

+

M N

++ minusminus

different fault directions Combined with empirical modedecomposition algorithm (EMD) the new travelling wavepolarity comparison protection principle based on the inte-gration of amplitude is derived It not only uses the initialtravelling wave but also uses the travelling wave after faulthappens So it is a reliable protection principle with obviousdirection discrimination By the way this new protectionprinciple is not affected by different initial angles differentgrounding resistance different bus structures and differentfault locations To verify the characteristics of the new prin-ciple a simulation based on PSCADEMTDC is carried onAnd the PSCAD simulation proved that this new protectionprinciple has the characteristics mentioned above indeed

The high operation speed is a very important advantagefor travelling wave protection Considering the differentparts of the new travelling waversquos operation time the newprotection principle can determine if the fault is internal orexternal in 5ms Then it can send the signal to breaker tooperate So it can be called an ultra-high-speed travellingwave protection

2 Traditional Travelling Wave PolarityComparison Protection

The direction element of the protection principle is a polaritycomparison relay It detects the initial voltage travelling waveand current travelling wave as comparison objects Whenthe voltage travelling wave and current travelling wave haveopposite polarities of both sides the internal fault can bedetermined When the voltage travelling wave and currenttravelling wave have same polarities of any side the externalfault can be determinedThe schematic of protection is shownin Table 1 (Superimposed voltage in the table appears at themoment that fault happens It has same value and oppositepolarity with the voltage on the transmission line just beforethe fault moment And it is the voltage source in the circuit ofthe table indeed)

3 Empirical Mode Decomposition

The empirical mode decomposition algorithm can distin-guish the different scale fluctuations or trends in the signalgradually And the result of EMD is a series of differentcharacteristic scales data called intrinsic mode functions(IMF) [17ndash19]

The result of EMD can be described as

119878 (119905) =

119873

sum

119894=1

IMF119894 (119905) + 119877 (119905)

(1)

where 119878(119905) is the EMD result IMF119894(119905) is 119894 order intrinsicmode

function and 119877(119905) is trends signalIntrinsic mode function is a single component signal and

it must meet the following two conditions (1) differencebetween the number of extreme points and zero crossingpoints is not more than one over the entire length of thesignal (2) the envelope of IMF is symmetry about time axis

The processes of EMD are described as the followingsteps

(1) Find all the maxima of the original signal 119878(119905) Thenthe maxima envelope 119864

+(119905) can be calculated based on cubic

spline interpolation Similarly the minimum envelope 119864

minus(119905)

can also be calculated Then the average envelope 119872(119905) canbe defined as

119872(119905) =

119864

+ (119905) + 119864

minus (119905)

2

(2)

(2) Let 119878(119905) be minus119872(119905) to get a new signal11986711(119905)

119867

1

1(119905) = 119878 (119905) minus 119872 (119905)

(3)

Then check if the following condition can be met

sum[119867

119896

1(119905) minus 119867

119896minus1

1(119905)]

2

sum[119867

119896minus1

1(119905)]

2le 120576

(4)




1(119905) can be

defined as

IMF1 (

119905) = 119867

119896

1(119905)

(5)


signal 119877(119905) as

119877 (119905) = 119878 (119905) minus IMF1 (

119905) (6)










119894

0= limΔ119909rarr0

Δ119876

Δ119905

= limΔ119909rarr0

119862119906

0Δ119909

Δ119905

= 119862119906

0V (7)



119864 = limΔ119909rarr0

ΔΦ

Δ119905

= limΔ119909rarr0

119871119862119906

0VΔ119909

Δ119905

= 119871119862119906

0V2 (8)



0 Then the

wave speed will be

V =

1

radic

119871119862

(9)

minus500

50

um (k

V)

im (k

A)

umim

f (kV

)im

imf (

kA)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

minus04minus02

0

minus010

01

minus010

01


Bus

Bus F

F

R1R2

HH

E E

i0i0

C2C9984002

u0

Δx

C1

fminus

f+

f0

f998400+



119906

0

119894

0

=

radic

119871

119862

(10)



119896 =

|119894|

|119906|

(11)



120582 =

int

119905119865+119889

119905=119905119865

119896119906 (119905) 119894 (119905) 119889119905

int

119905119865+119889

119905=119905119865

119894

2(119905) 119889119905

(12)


where 119905




120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

(13)



119906 = 119906

++ 119906

minus= (1 + 119896

119906119891) 119906

minus

119894 = 119894

++ 119894

minus= (1 + 119896

119894119891) 119894

minus

(14)



minus



119906119891


119894119891is the


119906119891| |119896

119894119891| le 1


119896 =

|119894 (119905)|

|119906 (119905)|

=

(1 + 119896

119894119891)

(1 + 119896

119906119891)

1003816

1003816

1003816

1003816

119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119906

minus (119905)

1003816

1003816

1003816

1003816

(15)


119896 = minus

(1 + 119896

119894119891)

(1 + 119896

119906119891)

119894

minus (119905)

119906

minus (119905)

(16)

Take (14) and (16) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= minus1 (17)



119906 = 119906

1015840

+= 119896

119906119911119906

minus

119894 = 119894

1015840

+= 119896

119894119911119894

minus

(18)




minusis the

minus1 +10








119906119911 119896

119894119911ge 0


119896 =

|119894 (119905)|

|119906 (119905)|

=

1003816

1003816

1003816

1003816

119896

119894119911119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119896

119906119911119906

minus (119905)

1003816

1003816

1003816

1003816

(19)


119896 =

|119894 (119905)|

|119906 (119905)|

=

119896

119894119911

119896

119906119911

119894

minus (119905)

119906

minus (119905)

(20)

Take (18) and (20) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= 1 (21)










Yes

Yes

No

No

another end

from another end

Return

Internal fault

External fault

Return

External fault

Breaker action

of time

Return




0 lt 120582 le 1

minus1 le 120582998400 lt 0




119877 119883 119866 119861









M NR1 R2F1F2 F3

CSCSCSCS

L2 = 100 km L1 = 200 km L3 = 100 km


minus005

0

005

ku

(kV

)

minus005

0

005

i (kA

)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)







53 Relater Factors






(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01












M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3


M NR1 R2F

CS CS

L2L1

tftm

tn





119891 119905119898 and 119905



Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)



















F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1(119905) can be

defined as

IMF1 (

119905) = 119867

119896

1(119905)

(5)


signal 119877(119905) as

119877 (119905) = 119878 (119905) minus IMF1 (

119905) (6)










119894

0= limΔ119909rarr0

Δ119876

Δ119905

= limΔ119909rarr0

119862119906

0Δ119909

Δ119905

= 119862119906

0V (7)



119864 = limΔ119909rarr0

ΔΦ

Δ119905

= limΔ119909rarr0

119871119862119906

0VΔ119909

Δ119905

= 119871119862119906

0V2 (8)



0 Then the

wave speed will be

V =

1

radic

119871119862

(9)

minus500

50

um (k

V)

im (k

A)

umim

f (kV

)im

imf (

kA)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

20 40 60 80 1000t (120583s)

minus04minus02

0

minus010

01

minus010

01


Bus

Bus F

F

R1R2

HH

E E

i0i0

C2C9984002

u0

Δx

C1

fminus

f+

f0

f998400+



119906

0

119894

0

=

radic

119871

119862

(10)



119896 =

|119894|

|119906|

(11)



120582 =

int

119905119865+119889

119905=119905119865

119896119906 (119905) 119894 (119905) 119889119905

int

119905119865+119889

119905=119905119865

119894

2(119905) 119889119905

(12)


where 119905




120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

(13)



119906 = 119906

++ 119906

minus= (1 + 119896

119906119891) 119906

minus

119894 = 119894

++ 119894

minus= (1 + 119896

119894119891) 119894

minus

(14)



minus



119906119891


119894119891is the


119906119891| |119896

119894119891| le 1


119896 =

|119894 (119905)|

|119906 (119905)|

=

(1 + 119896

119894119891)

(1 + 119896

119906119891)

1003816

1003816

1003816

1003816

119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119906

minus (119905)

1003816

1003816

1003816

1003816

(15)


119896 = minus

(1 + 119896

119894119891)

(1 + 119896

119906119891)

119894

minus (119905)

119906

minus (119905)

(16)

Take (14) and (16) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= minus1 (17)



119906 = 119906

1015840

+= 119896

119906119911119906

minus

119894 = 119894

1015840

+= 119896

119894119911119894

minus

(18)




minusis the

minus1 +10








119906119911 119896

119894119911ge 0


119896 =

|119894 (119905)|

|119906 (119905)|

=

1003816

1003816

1003816

1003816

119896

119894119911119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119896

119906119911119906

minus (119905)

1003816

1003816

1003816

1003816

(19)


119896 =

|119894 (119905)|

|119906 (119905)|

=

119896

119894119911

119896

119906119911

119894

minus (119905)

119906

minus (119905)

(20)

Take (18) and (20) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= 1 (21)










Yes

Yes

No

No

another end

from another end

Return

Internal fault

External fault

Return

External fault

Breaker action

of time

Return




0 lt 120582 le 1

minus1 le 120582998400 lt 0




119877 119883 119866 119861









M NR1 R2F1F2 F3

CSCSCSCS

L2 = 100 km L1 = 200 km L3 = 100 km


minus005

0

005

ku

(kV

)

minus005

0

005

i (kA

)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)







53 Relater Factors






(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01












M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3


M NR1 R2F

CS CS

L2L1

tftm

tn





119891 119905119898 and 119905



Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)



















F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where 119905




120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

(13)



119906 = 119906

++ 119906

minus= (1 + 119896

119906119891) 119906

minus

119894 = 119894

++ 119894

minus= (1 + 119896

119894119891) 119894

minus

(14)



minus



119906119891


119894119891is the


119906119891| |119896

119894119891| le 1


119896 =

|119894 (119905)|

|119906 (119905)|

=

(1 + 119896

119894119891)

(1 + 119896

119906119891)

1003816

1003816

1003816

1003816

119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119906

minus (119905)

1003816

1003816

1003816

1003816

(15)


119896 = minus

(1 + 119896

119894119891)

(1 + 119896

119906119891)

119894

minus (119905)

119906

minus (119905)

(16)

Take (14) and (16) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= minus1 (17)



119906 = 119906

1015840

+= 119896

119906119911119906

minus

119894 = 119894

1015840

+= 119896

119894119911119894

minus

(18)




minusis the

minus1 +10








119906119911 119896

119894119911ge 0


119896 =

|119894 (119905)|

|119906 (119905)|

=

1003816

1003816

1003816

1003816

119896

119894119911119894

minus (119905)

1003816

1003816

1003816

1003816

1003816

1003816

1003816

1003816

119896

119906119911119906

minus (119905)

1003816

1003816

1003816

1003816

(19)


119896 =

|119894 (119905)|

|119906 (119905)|

=

119896

119894119911

119896

119906119911

119894

minus (119905)

119906

minus (119905)

(20)

Take (18) and (20) to (13)

120582 =

sum

119905119865+119889

119905=119905119865119896119906 (119905) 119894 (119905)

sum

119905119865+119889

119905=119905119865119894

2(119905)

= 1 (21)










Yes

Yes

No

No

another end

from another end

Return

Internal fault

External fault

Return

External fault

Breaker action

of time

Return




0 lt 120582 le 1

minus1 le 120582998400 lt 0




119877 119883 119866 119861









M NR1 R2F1F2 F3

CSCSCSCS

L2 = 100 km L1 = 200 km L3 = 100 km


minus005

0

005

ku

(kV

)

minus005

0

005

i (kA

)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)







53 Relater Factors






(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01












M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3


M NR1 R2F

CS CS

L2L1

tftm

tn





119891 119905119898 and 119905



Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)



















F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Yes

Yes

No

No

another end

from another end

Return

Internal fault

External fault

Return

External fault

Breaker action

of time

Return




0 lt 120582 le 1

minus1 le 120582998400 lt 0




119877 119883 119866 119861









M NR1 R2F1F2 F3

CSCSCSCS

L2 = 100 km L1 = 200 km L3 = 100 km


minus005

0

005

ku

(kV

)

minus005

0

005

i (kA

)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)







53 Relater Factors






(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01












M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3


M NR1 R2F

CS CS

L2L1

tftm

tn





119891 119905119898 and 119905



Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)



















F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










(kV

)i (

kA)

20 40 60 80 1000Time (120583s)

20 40 60 80 1000Time (120583s)

minus01

0

01

minus01

0

01












M NR1 R2F1

F2

L2 = 100 km L1 = 200 km L3 = 100 km

CS CS CS CS

F3


M NR1 R2F

CS CS

L2L1

tftm

tn





119891 119905119898 and 119905



Δ119905 = 119905

119898minus 119905

119899=

119871

1minus 119871

2

V

(22)



















F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























F1


F2










M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













M NR1 R2F

L

L2L1

CS CS

120582m 120582n



7 Conclusion






Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article ultra-high-speed travelling wave...

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